flux of $\langle x,y,z^2\rangle$ across unit sphere

Question

I'm trying to compute $\iint_S F\cdot$ dS where $F=\langle x,y,z^2\rangle$ and $S$ is the unit sphere centered at the origin.

Here's my attempt:

On the sphere we can describe any point by $r(\phi,\theta)=\langle \sin\phi \cos\theta,\sin\phi \cos\theta, \cos\theta\rangle$ and the outward normal vector to $S$ is given by $n=r.$ Therefore,

$$\iint_S F\cdot dS = \iint_D F(r(\phi,\theta))\cdot n dA$$ $$=\iint_D \langle \sin\phi\cos\theta,\sin\phi\sin\theta,\cos^2\theta\rangle\cdot \langle \sin\phi\cos\theta,\sin\phi\sin\theta,\cos\theta\rangle dA$$ $$=\iint_D \sin^2\phi(\cos^2\theta + \sin^2\theta)+\cos^3d\theta dA = \int_0^{\pi}\int_0^{2\pi} \sin^2\phi +\cos^3\theta d\theta d\phi = \pi^2$$

however the answer is $\frac{8}{3}\pi$ and the last equals sign is correct (used Wolfram to confirm), so I guess I did something wrong in the setup.

Answer 1

The position vector should be $$r(\phi,\theta)=\langle \sin\theta \cos\phi,\sin\theta \sin\phi, \cos\theta\rangle.$$ And the volume element should be $$ dA = \langle \sin\theta \cos\phi,\sin\theta \sin\phi, \cos\theta\rangle \sin \theta \ d\theta \ d\phi.$$ So the integral should be $$\iint_S F\cdot dA = \int_{\theta = 0}^{\theta = \pi} \int_{\phi = 0}^{\phi = 2\pi} \left( \sin^2 \theta + \cos^3 \theta\right) \sin \theta \ d\theta \ d\phi.$$

Answer 2

You can use the divergence theorem to simplify things. We have $$\operatorname{div} \vec{F}(x,y,z) = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} = 2+2z$$

so $$I = \iint\limits_{\text{unit sphere}} \vec{F}\cdot d\mathbf{S} = \iiint\limits_{\text{unit ball}} \operatorname{div} \vec{F} \,dV = \int_{\phi = 0}^{2\pi} \int_{\theta = 0}^\pi \int_{r=0}^1 2(1+r\cos\theta)\underbrace{r^2\sin\theta\,dr\,d\theta\,d\phi}_{dV}$$

The integrand is independent of $\phi$ so it factors out as $2\pi$, and then the second integral vanishes:

$$I = 4\pi\left(\int_{\theta = 0}^\pi \int_{r=0}^1 r^2\sin\theta\,dr\,d\theta + \int_{\theta = 0}^\pi \int_{r=0}^1 r^3\cos\theta\sin\theta\,dr\,d\theta\right) = \frac83\pi$$